=>(x-2001)²\(\le\frac{49}{12}\approx4,08\)

=>(x-2001)²={0;1;4}

TH1: (x-2001)²=0

=>x=2001

=>y=7

TH2: (x-2001)²=1

\(\Rightarrow\hept{\begin{cases}x-2001=1\\x-2001=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=2002\\x=2000\end{cases}}\)

=>y²=37(loại)

TH3: (x-2001)²=4

\(\Rightarrow\hept{\begin{cases}x-2001=2\\x-2001=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=2003\\x=1999\end{cases}}\)

=>y²=1

=>y=1

 Vậy (x;y)=(2001;7);(2003;1);(1999:1)

bài này có lộn đề không thế

ta có: 49 - y² = 12(x - 2001)²

=> \(12\left(x-2001\right)^2\le49\\ \Rightarrow\left(x-2001\right)^2\le\frac{49}{12}\approx4\)

mà (x - 2001)² là số chính phương

=> \(\left(x-2001\right)^2=\left\{0;1;4\right\}\)

nếu (x - 2001)² = 0

=> x - 2001 = 0 => x = 2001

=> 49 - y² = 0 => y² = 49 \(\Rightarrow\left[\begin{matrix}y=7\\y=-7\left(loại\right)\end{matrix}\right.\)

nếu (x - 2001)² = 1

\(\left\{\begin{matrix}x-2001=1\\x-2001=-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2002\\x=2000\end{matrix}\right.\)

\(\Rightarrow49-y^2=12\Rightarrow y^2=37\left(loại\right)\)

nếu (x - 2001)² = 4

\(\Rightarrow\left\{\begin{matrix}x-2001=2\\x-2001=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2003\\x=1999\end{matrix}\right.\)

\(\Rightarrow49-y^2=12.4=48\Rightarrow y^2=1\Rightarrow\left\{\begin{matrix}y=1\\y=-1\left(loại\right)\end{matrix}\right.\)

vậy ta có các cặp (x;y) là (2001;7), (2003;1), (1999;1)

2^x+1.3^y=12^y

<=> 2^x+1.3^y=3^y.2^2y

<=> 2^x+1=2^2y

=> x+1=2y

\(2^{x+1}.3^y=12^x\)

\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)

\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)

\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)

Điều kiện đã cho \(\Leftrightarrow7\left(x-2019\right)^2+y^2=23\) (*)

Do \(\left(x-2019\right)^2,y^2\ge0\) nên (*) suy ra \(y^2\le23\Leftrightarrow y^2\in\left\{0,1,4,9,16\right\}\)

\(\Leftrightarrow y\in\left\{0,1,2,3,4\right\}\)

Hơn nữa, lại có \(y^2=23-7\left(x-2019\right)^2\). Ta thấy \(VP\) chia 7 dư 2.

\(\Rightarrow y^2\) chia 7 dư 2 \(\Rightarrow y\in\left\{3,4\right\}\)

Xét \(y=3\) \(\Rightarrow7\left(x-2019\right)^2=14\) \(\Leftrightarrow\left(x-2019\right)^2=2\), vô lí.

Xét \(y=4\Rightarrow7\left(x-2019\right)^2=7\) \(\Leftrightarrow\left(x-2019\right)^2=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=2018\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(4;2020\right),\left(4;2018\right)\right\}\) thỏa mãn ycbt.

\(\left(x-2016\right)^2\ge0\Rightarrow5-y^2\ge0\Rightarrow y^2\le5\Rightarrow y^2\in\left\{0,1,4\right\}\)

\(\Rightarrow5-y^2\in\left\{5,4,1\right\}\)Mà x,y là STN 

\(\Rightarrow5-y^2=4\Rightarrow y^2=1\Rightarrow y=1\)

\(\Rightarrow\left(x-2016\right)^2=4\Rightarrow\orbr{\begin{cases}x-2016=2\\x-2016=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=2018\\x=2014\end{cases}}}\)

Vậy ...

Hiêu hai cp=5 chỉ có 4 &9

​=>y=+-2; x-2016=+-3=>x=2019 hoac x=2013

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